pset3sol - 22.02 Problem Set 3 Solution 1.Solution The...

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Unformatted text preview: 22.02 Problem Set 3 Solution 1.Solution The energy eigenvalue problem for the given system is: H ˆ ψ = E ψ h 2 d 2 H ˆ = − 2 m dx 2 + V ( x ) | x | > a V ( x ) = − V | x | ≤ a For bounded particle, the solution is: ψ = Ae κ x x < − a 1 ψ 2 = B sin kx or B 'cos kx − a ≤ x ≤ a ψ 3 = Ce − κ x x > a k = 2 0 ) ( 2 h V E m + , κ = 2 2 h mE Since the solution has to be odd, ψ 2 = B sin kx Also we know, ψ 2 ( a ) = ψ 3 ( a ) , if ψ 3 ( x ) is decaying, there has to be at least a quarter of wave from 0 to a within the well. This means λ / 4 ≥ a . “Just being bounded” means λ / 4 = a and E = 0 . This gives us k = m a V mV a 2 2 2 0 2 0 8 2 4 2 2 h h π π λ π = ⇒ = = 2.Solution Liboff 5.12 (a) Proof [ A ˆ + B ˆ , C ˆ ] = [ A ˆ , C ˆ ] + [ B ˆ , C ˆ ] LHS = [ A ˆ + B ˆ , C ˆ ] = ( A ˆ + B ˆ ) C ˆ − C ˆ ( A ˆ + B ˆ ) = A ˆ C ˆ + B ˆ C ˆ − C ˆ A ˆ − C ˆ B ˆ RHS = [ A ˆ , C ˆ ] + [ B ˆ , C ˆ ] = A ˆ C ˆ − C ˆ A ˆ + B ˆ C ˆ − C ˆ B ˆ = LHS ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (b) Proof [ AB , C ] = A [ B , C ] + [ A , C ] B ˆ ˆ ˆ ˆ ˆ ˆ LHS = ABC − CAB RHS = A ˆ ( B ˆ C ˆ − C ˆ B ˆ ) + ( A ˆ C ˆ − C ˆ A ˆ ) B ˆ = A ˆ B ˆ C ˆ − A ˆ C ˆ B ˆ + A ˆ C ˆ B ˆ − C ˆ A ˆ B ˆ = A ˆ B ˆ C ˆ − C ˆ A ˆ B ˆ = LHS Liboff 5.19 ( ϕ 2 , ϕ 3 ) = {sin kx ,exp( ikx )} Since we know ϕ 3 = exp( ikx ) is a common eigenfunction of H ˆ and p ˆ , we’d like to keep it as one of the two common eigenfunctions composed of ( ϕ 2 , ϕ 3 ) . If we denote it as ϕ a , then ϕ a = a 2 ϕ 2 + a 3 ϕ 3 = ϕ 3 , which means a 2 = 0 and a 3 = 1 . (1) Also we know ϕ b = exp( − ikx ) is a common eigenfunction of H ˆ and p ˆ and is linearly independent of ϕ a = exp( ikx ) , we hope we could construct ϕ b as a linear combination of ( ϕ 2 , ϕ 3 ) . ϕ b = b 2 ϕ 2 + b 3 ϕ 3 = b 2 sin kx + b 3 exp( ikx ) = b 2 sin kx + b 3 (cos kx + i sin kx ) 22.02 Problem Set 3 Solution ϕ b = b 3 cos kx + ( b 2 + ib 3 )sin kx = exp( − ikx ) = cos kx − i sin kx ⇒ b 3 = 1, b 2 = − 2 i (2) Sum (1) and (2), we get: ϕ a = 0 × ϕ 2 + 1 × ϕ 3 = exp( ikx ) ϕ b = − 2 i × ϕ 2 + 1 × ϕ 3 = exp( − ikx ) ϕ 2 and ϕ 3 are a pair of linearly independent common eigenfunctions of H ˆ and p ˆ . Liboff 5.25 Use uncertainty principle: If [ A ˆ , B ˆ ] = C ≠ 0 , then ∆ A ∆ B ≥ 1 | < C > | 2 (a) ∆ x ∆ E [ x ˆ, E ˆ ] ψ = x ˆ E ˆ ψ − E ˆ x ˆ ψ h 2 d 2 x ˆ = x , E ˆ = − 2 m dx 2 + V ( x ) h 2 d 2 ψ h 2 d 2 [ x ˆ, E ˆ ] ψ = x [ − 2 m dx 2 + V ( x ) ψ ] − [ − 2 m dx 2 + V ( x )][ x ψ ] h 2 d 2 ψ h 2 d 2 = − x 2 m dx 2 + xV ( x ) ψ + 2 m dx 2 [ x ψ ] − xV ( x ) ψ h 2 d 2 ψ h 2 d d ψ h 2 d 2 ψ h 2 d ψ h 2 d 2 ψ h 2 d ψ + + + x + = − x 2 m dx 2 2 m dx [ ψ + x dx ] = − x 2 m dx 2 2 m dx 2 m dx 2 2 m dx [ x ˆ, E ˆ ] = h 2 d m dx d p ˆ x = − i h dx ⇒ [ x ˆ, E ˆ ] = i h p ˆ x m 1 h From uncertainty principle, we get ∆ x ∆...
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This note was uploaded on 11/09/2009 for the course PHY 4604 taught by Professor Mucciolo during the Spring '09 term at University of Central Florida.

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pset3sol - 22.02 Problem Set 3 Solution 1.Solution The...

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